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Math Help - Geometric Random Variable

  1. #1
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    Geometric Random Variable

    Let W be the number of failures until the first success; that is, W = T-1. Show that W has mean (1-pi)/pi and variance (1-pi)/(pi^2).

    I know for mean, E(T) = 1/pi and for variance Var(T) = (1-pi)/(pi^2).

    I guess I'm looking for E(T(W)) = E(T(T-1)) and Var(T(W)) = Var(T(T-1)).

    I've included how I've started so far in the attachment. Note I messed up on the last part in the step where I moved pi in front of summation. Should say (t-1). Sorry.

    Can someone help me with this proof.
    Attached Thumbnails Attached Thumbnails Geometric Random Variable-dsc06932.jpg  
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    Quote Originally Posted by DINOCALC09 View Post
    Let W be the number of failures until the first success; that is, W = T-1. Show that W has mean (1-pi)/pi and variance (1-pi)/(pi^2).

    I know for mean, E(T) = 1/pi and for variance Var(T) = (1-pi)/(pi^2).

    I guess I'm looking for E(T(W)) = E(T(T-1)) and Var(T(W)) = Var(T(T-1)).

    I've included how I've started so far in the attachment. Note I messed up on the last part in the step where I moved pi in front of summation. Should say (t-1). Sorry.

    Can someone help me with this proof.
    You will find the derivation of the mean and variance of a geometric random variable in most textbook on Mathematical Statistics. Here are some websites (found using Google in under 5 minutes):

    http://www.win.tue.nl/~rnunez/2DI30/...tributions.pdf

    Expectation and variance of the discrete geometric distribution. Digital explorations
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  3. #3
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    In my books it shows E[T] = 1/pi

    In the attachments it shows E[T] = 1/(1-pi)

    A little confused...

    PS, is my setp correct thus far?
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    Quote Originally Posted by DINOCALC09 View Post
    In my books it shows E[T] = 1/pi

    In the attachments it shows E[T] = 1/(1-pi)

    A little confused...

    PS, is my setp correct thus far?
    There are two different definitions for the geometric distribution: see Geometric distribution - Wikipedia, the free encyclopedia

    You have to decide which one is being used in your problem.
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    So...

    The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... } according to Wikipedia.

    Now, how do I change the summation in the next step to t=2 as the lower bounds. This is the step I'm stuck on.
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    Quote Originally Posted by DINOCALC09 View Post
    So...

    The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... } according to Wikipedia.

    Now, how do I change the summation in the next step to t=2 as the lower bounds. This is the step I'm stuck on.
    Did you read the links I gave you? The derivation is given.
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  7. #7
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    ok i get it now. will post my work once i'm done working it out on paper. thank you thank you thank you
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